Expert: Richard J. Raridon - 3/27/2008

Question
Hello,
 I need assistance with the following:
Find the zeros for the polynomial function and give the multiplicity for each zero.  State whether the graph crosses the x-axis or touches the x-axis and turns around, at each zero.

f(x)=1/5x^4 (x^2-5)(x-7)

I did the following:
5(1/5x^4)=0
x^2-5=0
x^2=5
x-7=0x=7

o, multiplicity 4, touches x-axis and turns around;7 multiplicity 1, crosses x-axis;5, multiplicity 2, touches x-axis and turns around.

Possible answers:
0, multiplicity 4, touches x-axis and turns around;7, multiplicity 1, crosses x-axis;square root of 5, multiplicity 1, crosses x-axis;square root of -5, multiplicity 1 crosses x -axis

0, multiplicity 4, touches x-axis and turns around;7 multiplicity 1, crosses x-axis;5, multiplicity 2, touches x-axis and turns around.

0, multiplicity 4, touches x-axis and turns around; 7, multiplicity 1, crosses x-axis

0, multiplicity 4, crosses x-axis; 7, multiplicity 1, touches x-axis and turns around; square root 5, multiplicity 1, touches x-axis and turns around; square root -5, multiplicity 1, touches x-axis and turns and turns around

Use the Intermediate Value Theorem to determine whether the polynomial function has a real zero between the given integers:

f(x)=2x^3+7x^2-10x-9, between -5 and -4

Possible answers:
f(-5)=-34 and f(-4)=-15; no
f(-5)=-34 and f(-4)=15;yes
f(-5)=34 and f(-4)=15;no
f(-5)=34 and f(-4)= -15;yes

Graph the polynomial function:

f(x)=1/3-1/3x^4

Find the zeros of the polynomial function:
f(x)=x^3-6x^2+9x

possible answers:
x=0,x=3
x=0,x=-3,x=3
x=1,x=3
x=0,x=-3

Use the graph or table to determine a solution of the equation. Use synthetic division to verify that this number is a solution of the equation.  Then solve the polynomial equation:

x^3+6x^2+11x+6=0

Possible answers:
-1;The remainder is 0;-1,-2, and 3 or {-2.-1,3}
-1; The remainder is 0,-1,2, and -3 or {-3,-1,-2}
-1, The remainder is 0, 1,-2, and -3 or {-3,-2,1}
-1; The remainder is 0, -1,-2, and -3 or {-3,-2,-1}




Determine whether the given quadratic function has a minimum value or maximum value. Then find the coordinates of the minimum or maximum point:

f(x)=-x^2-2x-5

I did the following:

f(x)=-x^2-2x-5
a=-1 b=-2 c=-5
a<0
x=-b/2a=-(-2)/2(-1)=2/-2=-1
f(x)=-x^2-2x-5 is f(-1)=-1(-1)^2-2(-1)-5--1+2-5=-4
I chose minimum (-1,-4)

The possible answers are:
maximum;-4,-1
maimum; -1,-4
minimum; -4,-1
minimum; -1,-4

I beleive I made an error and the correct answer should be maximum;-1,-4

Answer
For the first one, the correct answer is the first one after Possible answers.
2. You could have calculated f(-5)=-34 and f(-4=15
3. I can't graph a function for you.  Just put in a value for x, calculate f(x) and plot.  one point is (0,1/3)
4. f(x) = x(x-3)^2 so x = 0,3
5. x^3+6x^2+11x+6 = (x+1)(x+2)(x+3)
6. f(x)=x^2-2x-5 has a minimum at (1,-6)